# Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$

Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$

My attempt:

$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$

$$\frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}}$$

$$\frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}$$

$$\frac{(\sin(x)-\cos(x))}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}$$

$$\frac{1}{\sin(x)+\cos(x)}$$

Now this is where I am stuck , I thought of multiplying the numerator and denominator by $$\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}$$ but that did not work out well..

• I recommend starting by clearing the fractions in the question (ie multiply through by (sec x + csc x)(tan x - cot x). – almagest Mar 20 '16 at 19:49
• Proving that $\tan x + \cot x = \sec x\csc x$ and that $\sec^2x\csc^2x = \sec^2 x+\csc^2 x$ will go a long way in helping you prove this identity. – John Joy Mar 20 '16 at 21:30

multiply and divide $\frac{1}{\sin{x}+\cos{x}}$ by $\frac{1}{\sin{x}\cos{x}}$ then in the numerator substitute $1$ by $sin^2{x} + cos^2{x}$

• I don't really understand you mean by that , could you show me step by step please – bigfocalchord Mar 20 '16 at 20:02
• @dydxx What inquisitive means is that once you obtain $$\frac{1}{\sin x + \cos x}$$ you multiply its numerator and denominator by $\frac{1}{\sin x\cos x}$ to obtain \begin{align*} \frac{1}{\sin x + \cos x} & = \frac{1}{\sin x + \cos x} \cdot \frac{\frac{1}{\sin x\cos x}}{\frac{1}{\sin x\cos x}}\\ & = \frac{1 \cdot \frac{1}{\sin x\cos x}}{\frac{1}{\cos x} + \frac{1}{\sin x}} \end{align*} then replace the $1$ in the numerator by $\sin^2x + \cos^2x$ in order to obtain $$\frac{\frac{\sin^2x + \cos^2x}{\sin x\cos x}}{\frac{1}{\cos x} + \frac{1}{\sin x}}$$ Your proof should include equals signs. – N. F. Taussig Mar 20 '16 at 23:56

HINT:

$$\sec^2x-\tan^2x=1=\csc^2x-\cot^2x$$

$$\iff\sec^2x-\csc^2x=\tan^2x-\cot^2x$$

$$\sec^2x-\csc^2x=(\sec x-\csc x)(\sec x+\csc x)$$

Can you take it from here?

• This is the method I used and the easiest method. – N.S.JOHN Mar 21 '16 at 8:17

Take the right hand side:

$$\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}=\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)} + \frac{1}{\sin(x)}} = \frac{\sin^2x+\cos^2x}{\sin x+\cos x}=\frac{1}{\sin x+\cos x}$$

and this is equal to the expression you found for the left hand side.

• Ah okay I see , would there a way by just continuing where I left off and prove it equals the RHS – bigfocalchord Mar 20 '16 at 20:11
• @dydxx See the hint by inquisitive. – N. F. Taussig Mar 21 '16 at 0:05

Start by multiplying both sides by the denominators, so that you get $$\sec(x)^2-\csc(x)^2 = \tan(x)^2-\cot(x)^2.$$ Now starting from the left-hand side : \begin{align*}\sec(x)^2-\csc(x)^2 & =\frac{1}{\cos(x)^2}-\frac{1}{\sin(x)^2} \\ & = \frac{1}{\cos(x)^2}-1-\frac{1}{\sin(x)^2}+1 \\ & = \frac{1-\cos(x)^2}{\cos(x)^2}-\frac{1-\sin(x)^2}{\sin(x)^2} \\ & = \frac{\sin(x)^2}{\cos(x)^2}-\frac{\cos(x)^2}{\sin(x)^2}.\end{align*}

Multiply the LHS by product of the reciprocal of RHS, and the RHS. $$\begin{array}{lll} \displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}&=&\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}\cdot\frac{\sec x + \csc x}{\tan x + \cot x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\ &=&\displaystyle\frac{\sec^2 x - \csc^2 x}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\ &=&\displaystyle\frac{(\tan^2 x + 1) - (\cot^2 x+1)}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\ &=&\displaystyle\frac{\tan x + \cot x}{\sec x + \csc x}\\ \end{array}$$ The trick: if LHS=RHS then $LHS\times\frac{1}{RHS}$ will always equal 1, just as $\frac{1}{RHS}\times RHS$ will always equal 1.

Cross multiplying we need to prove that:

$$\sec ^2 x -\csc ^2 x = \tan ^2 x -\cot ^2 x$$

or to prove that

$$1+ \tan ^2 x -(1 +\cot ^2 x ) = \tan ^2 x -\cot ^2 x !!$$